Question

A particle executes SHM such that the maximum velocity during the oscillation is numerically equal to half the maximum acceleration. What is the time period?

Answer 1

A particle performs a linear S.H.M along a path 10 cm long. The particle starts from a distance of 1 cm from the mean position towards the positive extremity. Find the epoch and the phase of motion when the displacement is 2.5 cm.
Solution:
Given: Path length = 10 cm, amplitude = path length /2 = 10/2 = 5 cm, Initial displacement = x0 = 1 cm, Displacement = x = 2.5 cm.
To Find: Epoch = α =? and phase of S.H.M. = (ωt + α) = ?
Epoch = α = sin-1(xo/a) = sin-1(1/5) = sin-1(0.2) = 11°32’

Displacement of a particle performing S.H.M. is given by

Answer 2

Hey buddy,

● Answer-
T = 3.142 s

● Solution-
# Given-
vmax = 1/2 amax

# Solution-
We know that, for a particle in S.H.M.
vmax = Aω
amax = Aω^2

But we have,
vmax = 1/2 amax
Aω = 1/2 Aω^2
ω = 2 rad/s

Time period is given by
T = 2π/ω
T = 2π/2
T = π
T = 3.142 s

Hope that is useful...